Find the derivative of the functions from first principle: sin (x + 1)

Asked by Abhisek | 1 year ago |  79

##### Solution :-

Let f(x) = sin(x + 1). Accordingly, f(x + h) = sin(x + h + 1) By first principle,

$$f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$$

$$\lim\limits_{h \to 0}\dfrac{1}{h}[sin(x+h+1)-sin(x+1)]$$ $$\lim\limits_{h \to 0}\dfrac{1}{h}[2cos(\dfrac{2x+h+2}{2})sin(\dfrac{h}{2})]$$

$$\lim\limits_{h \to 0} [cos(\dfrac{2x+h+2}{2}).\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]$$

$$\lim\limits_{h \to 0} [cos(\dfrac{2x+h+2}{2}).\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]$$

$$\lim\limits_{h \to 0}\dfrac{1}{h} cos(\dfrac{2x+h+2}{2}). \lim\limits_{\dfrac{h}{2} \to 0}\dfrac{1}{h}\dfrac{sin(\dfrac{h}{2})}{(\dfrac{h}{2})}]$$

$$cos( \dfrac{2x+0+2}{2}).1$$

= cos(x+1)

Answered by Pragya Singh | 1 year ago

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